Answer
\[\int_{0}^{4\pi}\sqrt{5-4\cos t}\;\;dt\;\;\approx 26.7298\]
Work Step by Step
\[x=t-2\sin t\;,\;y=1-2\cos t\;\;\,0\leq t\leq 4\pi\]
Differentiate $x$ with respect to $t$:
\[\frac{dx}{dt}=1-2\cos t\]
Differentiate $y$ with respect to $t$:
\[\frac{dy}{dt}=2\sin t\]
Therefore the arc length is given by:
\[L=\int_{0}^{4\pi}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\]
\[L=\int_{0}^{4\pi}\sqrt{(1-2\cos t)^2+(2\sin t)^2}dt\]
\[L=\int_{0}^{4\pi}\sqrt{(1+4\cos^2 t-4\cos t)+(4\sin^2 t)}dt\]
\[L=\int_{0}^{4\pi}\sqrt{5-4\cos t}\;\;dt\]
By using calculator,
\[L\approx 26.7298\]
